It says here, in Malvino Electronic Principles, that the emitter voltage(AC) is half of the input voltage v1.

Why?

I understood the the current must be constant at this tail, which contains emitter resistor. Its constant because, there is constant voltage drop across that resistor, and 99% of the Vee is dropped across it, because 2 * 0.7 V of both base-emitter junctions, is negligible.

""It says here, in Malvino Electronic Principles, that the emitter voltage(AC) is half of the input voltage v1.

Why? ""

no nibbles yet? i've been thinking about it all day... here's my 'poor man's offering'...

you already know the necessary facts. So let's work it in your mind.

but first simplify it by removing Q2 as a step in our thinking...

with no Q2,,,
as input to Q1's base rises, more current flows through Q1 so voltage at emitter goes up.
In fact -emitter voltage will rise by almost exactly same amount as base did, because Vbe is pretty small. You already knew that.

now replace Q2 and observe how differently it works -

as input to Q1's base rises, more current flows through Q1 ,so the voltage at emitter still goes up but not quite like before.
It's different now because every extra millivolt at emitter is also impressed across Q2's e-b.
And in a direction to lower Q2's current which drives emitter voltage back down.

Qualitatively that's what is happening. And you knew that too.

To see why it's half Vin not 1/100th - let's try this thought...
allow for a moment this mild exaggeration: that the "tail" is a constant current source
(which it approximates if RsubE is high)

now do a Kirchoff's Voltage Law Walk around this loop, for the AC signal.
::
start at circuit common and traverse up through V1, in through Q1's b-e and out via Q2's e-b which gets us back to common.
-V1 +Vbe(Q1) + Veb(Q2) = 0
V1 = Vbe(Q1) + Veb(Q2)

That means for the AC signal the two base-emitter junctions form a voltage divider .

Aha - so the signal voltage will divide between them, and emitter node is in middle.
you knew that, too...

so far as AC signal current goes, the two base emitter junctions are in series. No ac signal current can get out of emitter -emitter node via a constant current source. Not much can get out through that RE if it's respectably high resistance, signal current will take the easier way out through Q2's internal Re.
you knew the facts leading to that.

So your AC signal voltage is divided between the internal Re's of the two transistors. For a differential amp those transistors would be a matched pair with similar internal resistances. THAT'S why it's half !

learning is mostly discovering what you already knew.....

Interesting - with Q2's base commoned, that differential amp amounts to just a common collector stage followed by a common base stage... common collector stage gives it current gain , common base gives voltage gain.

[ now at first glance it will look like i missed a sign in that KVL - remember this is small signal analysis not the DC biasing. ]

and yes,, op-amps are differential amplifiers. Don't despair you'll be amazed how quickly they make sense. They're not hard to understand, just hard to believe..
.

""It says here, in Malvino Electronic Principles, that the emitter voltage(AC) is half of the input voltage v1.

Why? ""

no nibbles yet? i've been thinking about it all day... here's my 'poor man's offering'...

you already know the necessary facts. So let's work it in your mind.

but first simplify it by removing Q2 as a step in our thinking...

with no Q2,,,
as input to Q1's base rises, more current flows through Q1 so voltage at emitter goes up.
In fact -emitter voltage will rise by almost exactly same amount as base did, because Vbe is pretty small. You already knew that.

now replace Q2 and observe how differently it works -

as input to Q1's base rises, more current flows through Q1 ,so the voltage at emitter still goes up but not quite like before.
It's different now because every extra millivolt at emitter is also impressed across Q2's e-b.
And in a direction to lower Q2's current which drives emitter voltage back down.

Qualitatively that's what is happening. And you knew that too.

To see why it's half Vin not 1/100th - let's try this thought...
allow for a moment this mild exaggeration: that the "tail" is a constant current source
(which it approximates if RsubE is high)

now do a Kirchoff's Voltage Law Walk around this loop, for the AC signal.
::
start at circuit common and traverse up through V1, in through Q1's b-e and out via Q2's e-b which gets us back to common.
-V1 +Vbe(Q1) + Veb(Q2) = 0
V1 = Vbe(Q1) + Veb(Q2)

That means for the AC signal the two base-emitter junctions form a voltage divider .

Aha - so the signal voltage will divide between them, and emitter node is in middle.
you knew that, too...

so far as AC signal current goes, the two base emitter junctions are in series. No ac signal current can get out of emitter -emitter node via a constant current source. Not much can get out through that RE if it's respectably high resistance, signal current will take the easier way out through Q2's internal Re.
you knew the facts leading to that.

So your AC signal voltage is divided between the internal Re's of the two transistors. For a differential amp those transistors would be a matched pair with similar internal resistances. THAT'S why it's half !

learning is mostly discovering what you already knew.....

Interesting - with Q2's base commoned, that differential amp amounts to just a common collector stage followed by a common base stage... common collector stage gives it current gain , common base gives voltage gain.

[ now at first glance it will look like i missed a sign in that KVL - remember this is small signal analysis not the DC biasing. ]

and yes,, op-amps are differential amplifiers. Don't despair you'll be amazed how quickly they make sense. They're not hard to understand, just hard to believe..
.

Thank you very much for your in-depth reply. I just finished reading it. I understood every word you said. Thank you very very much.

Differential amps are now clear. Now I can move on to op-amps, and nail them.

So, does the problem statement say the Rc = Re? Because the gain will only be 1/2 in that case.

This is the Vo/Vi gain. Bassalisk's question referred to Ve/Vi gain, which is 1/2 for matched transistors.

The common way to do this is with the T small-signal model of a NPN (see link below). Then do a KVL loop from Q1's base to Q2's base (this works because they are both ground referenced). Because there is only the AC source and two matched Re's in the loop the AC voltage at Ve is Vac/2.